题目内容

14.方程组$\left\{\begin{array}{l}x-y=3\\ 2x+y=6\end{array}\right.$的解为(  )
A.$\left\{\begin{array}{l}x=3\\ y=4\end{array}\right.$B.$\left\{\begin{array}{l}x=3\\ y=0\end{array}\right.$C.$\left\{\begin{array}{l}x=6\\ y=1\end{array}\right.$D.$\left\{\begin{array}{l}x=6\\ y=-1\end{array}\right.$

分析 利用①+②可消除y,从而可求出x,再把x的值代入①,易求出y,从而求解.

解答 解:$\left\{\begin{array}{l}{x-y=3①\\;}\\{2x+y=6②}\end{array}\right.$,
①+②,得3x=9,
解得x=3,
把x=3代入①,得
3-y=3,
解得y=0,
故原方程组的解是$\left\{\begin{array}{l}{x=3}\\{y=0}\end{array}\right.$.
故选:B.

点评 本题考查了解二元一次方程组,解题的关键是掌握加减法消元的思想.

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